Question

If ∆ABC is right angled at C, then the value of cos (A + B) is
A) 0
B) 1
C) 1/2
D) root3/2

Answer 1

Answer:

A) 0

Step-by-step explanation:

Since triangle ABC is right angled at C,

So, C = 90° and A+ B =90°

cos(A+B) = cos90°

= 0

Answer 2

Answer:

0

Step-by-step explanation:

We know that,

$\rm \angle A + \angle B + \angle C = 180^{\circ}$

$\rm \angle A + \angle B + 90^{\circ} = 180^{\circ}$

$\rm \angle A + \angle B= 180^{\circ} - 90^{\circ}$

$\rm \angle A + \angle B= 90^{\circ}$

$\rm \cos (A+B) = \cos (90^{\circ}) = 0$